Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
 
public class Main {
    public static double solve[][] = new double[3001][3001];
    public static double probabilities[] = new double[3000];
    public static void main(String args[])throws IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        long n=Long.parseLong(br.readLine());
        String S[]=br.readLine().split(" ");
        for(int i=0;i<n;i++)
            probabilities[i]=Double.parseDouble(S[i]);
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
                solve[i][j]=0;
        }
        solve[0][0]=1;
        for(int i=0;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                solve[i][j+1]+=solve[i][j]*(1-probabilities[j]);
                solve[i+1][j+1]=solve[i][j]*(probabilities[j]);
            }
        }
        double totalProbability=0.0d;
        for(int i=0;i<(n+1)/2;i++)
            totalProbability+=solve[i][(int)n];
        System.out.println(1-totalProbability);
 
    }
}